Permutations: Examples

This lesson covers a few examples relating to permutations, particularly those involving \( {}^{n}P_r \).

There isn’t be anything new to be covered here. The problems can be solved using the methods we’ve already seen. This is just to get you familiarised with factorials and notations.

Example 1 Find the number of 4-letter sequences that can be formed using the English alphabets, such that alphabets are not repeated in any sequence.

Solution This equals the number of permutations of 26 different objects, taken 4 at a time – \( {}^{26}P_4 \). We could also have computed the answer using the multiplication principle (as we did previously). The answer will come out to be 26 x 25 x 24 x 23 (which is same as \( {}^{26}P_4 \))

Example 2 How many 4-digit numbers can be formed using the digits 3, 4, 5 and 7, using each digit only once?

Solution The number of numbers in this case would equal \( {}^{4}P_4 \) or 4! – number of permutations of 4 different objects, taken all at a time.

Let’s complicate things a bit.

Example 3 Find the number of 3-digit numbers formed using the digits 1 to 9, without repetition, such the numbers either have all digits less than 5 or all digits greater than 4.

Solution Here, 342 is a possible number, 769 is another, but 385 is not. We can divide the numbers to be counted into two cases.

For those which have all the digits less than 5, we have to count the number of permutations of 4 digits (1, 2, 3, and 4), taken three at a time: \( {}^{4}P_3 \).

For those which have all the digits greater than 4, we have to count the number of permutations of 5 digits (5, 6, 7, 8 and 9), taken three at a time: \( {}^{5}P_3 \).

The total number of numbers would be \( {}^{4}P_3 + {}^{5}P_3 \), which equals 24 + 20 = 44.

That’s it for now. In the next lesson, I’ll talk about circular arrangements of objects.